Five persons \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}\) and \(\mathrm{E}\) are seated in a circular arrangement. If each of them is given a cap of one of the three colours red, blue and green, then the number of ways of distributing the caps such that the persons seated in adjacent seats get different coloured caps, is

There are 5 caps and 3 colours.
\(\therefore \) At least one colour will get repeated. As adjacent caps should be of different colours, no colour can repeat thrice.
\(\therefore \) Exactly two colours will repeat twice.
\(\therefore \) Colour of the caps are selected in 3 ways as follows:
Red-Red-Green-Green-Blue, Red-Red-Green-Blue-Blue, Red-Green-Green-Blue-Blue.
Now, while distributing the caps from above combinations, we choose any one of the 5 persons and give single colour cap. And remaining four caps can be distributed in alternate colour sequence, clock-wise or anticlock-wise.
This can be done in \(5 \times 2\) ways.
\(\therefore \) Required number of ways \(=3 \times 5 \times 2=30\).